SUMO-2

Let us solve for $x$ in terms of $y, z$ to yield:

$(yz + 4 -2y - 2z)x = 2yz + 7 - 4y - 4z$ ;

or $x = \frac{2yz + 7 - 4(y+z)}{yz + 4 - 2(y+z)}$ ;

or $x = \frac{2y(z-2) - 4(z-2) - 1}{(z-2)y - 2(z-2)}$ ;

or $x = \frac{2(y-2)(z-2) - 1}{(y-2)(z-2)}$ ;

or $x = 2 - \frac{1}{(y-2)(z-2)}$ .

In order for $x,y,z \in \mathbb{Z}$ , we require that the denominator be set to $(y-2)(z-2) = \pm 1$ . This can only be accomplished for the pairs:

$(y,z) = (1,1); (3,3); (1,3); (3,1)$ :

which finally yield the FOUR ordered-triplets:

$\boxed{(x,y,z) = (1,1,1); (1,3,3); (3,1,3); (3,3,1)}$

A simple use of SFFT!!!

Rearranging and factoring yields $(x-2)(y-2)(z-2)=1$ , which has 4 integer solutions.